TITLE: $R$ be a commutative ring with unity satisfying a.c.c. on radical ideals ; is it true that $R[x]$ also satisfies a.c.c. on radical ideals ?
QUESTION [5 upvotes]: $R$ be a commutative ring with unity satisfying ascending chain condition on radical ideals ; is it true that $R[x]$ also satisfies  ascending chain condition on radical ideals ?

REPLY [4 votes]: This is true and is a theorem of Ohm and Pendleton (Theorem 2.5 of this paper).  Here's a sketch of the proof.  Say that an ideal is radically finitely generated if its radical is the radical of a finitely generated ideal.  The acc on radical ideals is equivalent to every ideal being radically finitely generated.  An ideal which is maximal among the non-radically finitely generated ideals can be shown to be prime.  Taking such a maximal counterexample $P$ in $R[x]$, $P\cap R$ is radically finitely generated by hypothesis, and so after modding out $P\cap R$, $P$ will still not be radically finitely generated.  We can thus assume $R$ is a domain and $P\cap R=0$.  Now let $K$ be the field of fractions of $R$ and note that $PK[x]$ can be generated by a single polynomial $f\in P$.  If $c$ is the leading coefficient of $f$, then $P+(c)$ is radically finitely generated by maximality of $P$.  You can then show that $P$ is the radical of the ideal generated by $f$ and the $P$-components of the elements of $P+(c)$ that radically generate it.  Thus $P$ is radically finitely generated, which is a contradiction.